chemistry 2Is that rate = k[AB]^2? And I assume the units are L/mol*s. If so then it is a second order reaction and (1/A) - (1/Ao) = kt Substitute and solve for A
February 28, 2015

scienceTakes the object 6s to get where? 60m/6s = 10 m/s and since it's going up I would put a - sign on it.
February 28, 2015

ChemA is not true. You would LIKE about 10^5 difference in k1 and k2 but usually we can live with 10^3(1000 times) and in some cases even 10^2 (100 time) but never as close as 10^1 (10 times) I don't believe B is true but the statement isn't really all that clear to me. C ...
February 27, 2015

ChemistryThe problem doesn't ask for pH. I would redo it as a quadratic. 0.316 = (H^+) with the assumption that 2.0-x = 2.0 But it really is 2.0-x and it doesn't appear to me that the x can be ignored with respect to 2.0. That looks like about 15%. I would feel better if the ...
February 27, 2015

Chemistry...........H3PO3 ==> H^+ + H2PO3^- I..........2.0.......0......0 C...........-x.......x......x E..........2.0-x.....x......x Ignore the contribution to H^+ of k2 then substitute the E line into k1 expression and solve for x = (H^+)=(H2PO3^-)
February 27, 2015

ChemistryI got x^2 + 6.25E-3 - 2.06E-4 = 0 and when I solved that I came up with 0.0116M (vs 0.0144 before) for x. That gave me 1.93 for pOH and 12.06 for pH.
February 27, 2015

ChemistryYou have assumed 0.033-x = 0.033 and I'm not sure you can do that. If we compare the answer of 0.0144M for OH^- (rounded) is with the assumption but 0.033-0.0144 appears it isn't = 0.033. So I think you need to go through the quadratic equation. A quick guess is that ...
February 27, 2015

ChemistryI think Kb for the ascorbate ion is kw/k2 instead of kw/k1. Everything else (the procedure that is) looks ok to me. You can prove this is you wish by writing the k1 and k2 expression as well as Keq for the hyrolysis equation for ascorbate ion.
February 27, 2015

Chemistry HelpBa(OH)2(aq) + 2HBr(aq) → BaBr2(aq) + 2H2O(l) By the way, I see I made a typo late last night. I said your 0.0759 mols HBr was right; I should have said your 0.00759 mols HBr. Here is how you get the 1/2. Use the coefficients in the balanced equation. 0.00759 mols HBr x (...
February 27, 2015

Chemistry HelpYou are right with 0.0759. That is mols HBr. Then mols Ba(OH)2 is 1/2 that and not 2x that. Finally, the third is an error also. It should be M Ba(OH)2 = mols Ba(OH)2/L Ba(OH)2. You know mols and M, solve L and convert to mL.
February 27, 2015

ChemistryRemember that you can convert any material in mols in the equation to any other material in mols by using the coefficients in the balanced equation. 0.65 mole NH3 x (1 mol N2/2 mol NH3) = 0.65 x 1/2 = 0.325 mols N2 reacted and I know that's too many s.f. Important: Note ...
February 27, 2015

ChemistryUse the periodic table. There are 5 elements that are gases; H2, N2, O2, F2. Cl2 (plus of course the noble gases but we usually ignor those anyway). There are some common compounds that yu will learn as you go along; i.e., CO2, SO2, SO3, CO, NO, NO2, CH4(methane), C2H6(ethane...
February 27, 2015

ChemistryThere are 6.02E23 atoms in a mole of atoms or 6.02E23 molecules in a mole o molecules; actually, 6.02E23 anythings in a mole of anythings. Just as there are 12 in a DOZEN, 144 in a GROSS, 500 sheets in a REAM of paper,etc, there are 6.02E23 molecules in a mole. It's just ...
February 27, 2015

quadratic formulaFirst you should factor out the 2 to leave X^2 + 4x -5 = 0. You should not be getting an i. Getting an i means you are a negative under the square root. Your problem is that you're not doing the part under the square root sign correctly. sqrt(b^2-4*a*c) sqrt(16-4*1*-5) ...
February 27, 2015

Chemistry Need HelpThis is a limiting reagent problem (LR) and you know that because amounts are given for BOTH reactants. Your 1.50 mols Cl2 will produce 2*1.50 = 3.00 mols ClF if all of the Cl2 reacts and you have an excess of F2. Your 1.75 mols F2 will produce 2*1.75 = 3.50 mols ClF if you ...
February 27, 2015

ChemistrypH = pKa + log (base)/(acid) 5.25 = pKa + log (base)/(M x L) Solve for base and since the acid is in mols, the base will be in mols. Then mols = grams/molar mass. You know molar mass and mols, solve for grams. That will be the number of grams of the salt to add to the 532.9 mL...
February 27, 2015

Chemistry HelpNo, if the problem states it correctly, the theoretical yield is 12.2g. You collected only 8.82 g so %yield = (8.82/12.2)*100 =
February 27, 2015

chemistrymols PbCl2 = grams/molar mass = approx 6E-3 but you need a better number than that. I've estimated as well as all of the calculations that follow. That's mols in 2.00 mL out of 84.8 mL. Convert to mols in 100 mL (the original volume) by 6E-3 x (100/84.8) = approx 7E-3 ...
February 26, 2015

chemistryA quickie fix. That is Cu(OH)2. I'll try. You are right that there is no reaction. When you talk about the activity series you are talking about single replacement reactions. This is not a SR reaction so none of what you suggested applies. That takes care of that. When ...
February 26, 2015

scienceThe hot tea cools, the spoon and air heat up but the spoon heats more than the room full of air. The final temperature will be somewhat less than 65 F but not much less (it all depends upon the size of the room). You could calculate the final temperature if you had the masses ...
February 26, 2015

Chem, Oxidation StatesRule 1 is that the oxidation states must add to zero for a compound or to the charge on an ion. Rule 2 is representative elements have their "normal" charge; i.e., group I has +1, group II has +3, group III has +3, etc. Rule 3 is that H is +1 (except in unusual cases...
February 26, 2015

Chemistry, HELP!X(H2MoO4) = (H^+)^2/D X(HMoO4^-) = k1(H^+)/D X(MoO4^2-) = k1k2/D where D is denominator and that = (H^+)^2 + k1(H^+) + k1k2 when you finish calculating each mole fraction, then X*5.1E-4 = (that specie) There is another way to do this without mole fractions. It's about as ...
February 26, 2015

Chemistry EquilibriumPlease look at your post and your problem. If ALL of the 1.15 g H2O decomposed to 2H2O --> 2H2 + O2 you would have (1.15g H2O x (1 mol H2O/18 g H2O) x (1 mol O2/2 mol H2O) = about 0.064/2 = about 0.032 mols O2 formed. The problem states that 0.15 mol O2 were found at ...
February 26, 2015

Chemistry to DrBob222Thank you but I really don't deserve any credit. All of the homework helpers on this site are volunteers (no pay in dollars but much pay in satisfaction) and we do this because we like to do it. Most of us are retired teachers. I taught at a university for almost 40 years ...
February 26, 2015

ChemistryOk, Need some more help... This is what I have so far. 4. Urea (NH2)2CO is prepared by reacting ammonia with carbon dioxide. The byproduct is water. 637.2g of ammonia are reacted with 787.3 g of carbon dioxide. molar mass:NH3=17 CO2=44 CO(NH2)2=60.1 a. (10 points) Write a ...
February 26, 2015

Chemistry Drbob222a. (10 points) Calculate the theoretical yield of Ti(s). In order to find TY I have to find the limiting agent.. ?? ugh.. Ok, Ti=47.88 Cl4=35.45*4... To find moles in TiCl4 I take 3.54*10^7 * 1moleTiCl4/189.68 = 184,521.89 moles? But I need TY of Ti?? 7.91*10^6 * 1mole Ti/47....
February 26, 2015

Math Help!None of those numbers fit. Do you have a picture of the cylinder? Is it a regular (right) cylinder or an ellipsoid cylinder. It must not be a right cylinder. Here is a picture of a right cylinder. You can google image cylinder and see what the ellipsoid one looks like.
February 25, 2015

Math Help!volume of large cylinder = pi*r^2*h = 3.14(24)^2(18) = ? cc. For the small cylinder V = pi*r^2*h ?cc from above = 3.14(4)^2*h Solve for volume of large cylinder, substitute that for volume of small cylinder and solve for h of small cylinder.
February 25, 2015

Math Help!Is the volume of the small cylinder the same as the volume of the large cylinder. I think it must be; otherwise, I don't believe you can solve the problem.
February 25, 2015

ChemistryThe reason you have a negative sign is because you dropped the sign on the -0.8. When that term is moved to the right it changes to a + value. I don't know why you came up with 187. You didn't show your work so I can't find the error but I didn't end up with ...
February 25, 2015

chemistryI believe this is a two equation to be solved simultaneously problem. I suppose you want mass % NaCl IN THE SAMPLE. Let x = g NaCl and y = g sucrose ================== eqn 1 is x + y = 10.20 pi = iMRT. Substitute pi, i, R, and T and solve for M. Remember i for NaCl is 2 and ...
February 25, 2015

ScienceThe molar mass of dry air is about 29 g/mol. Since water has a smaller molar mass than air (29 for air; 18 for water), the density of humid aid is less than that of dry air so the weight of air/given volume is less.
February 25, 2015

Quick chemistry check1 is ok. 2 is ok but if your prof is picky about the number of significant figures you may want to adjust this. 3. Is that 35.0 GRAMS. If so I get the 2.15 but I don't agree with the exponent. 4 same comment about s.f. 5 I used 74 x 5.6 and didn't get 318 6. I used 24....
February 25, 2015

chemistrysure. Since it is an endothermic reaction, increasing T will shift it to the right and that will produce more NO
February 25, 2015

chemistryN2 + O2 + heat ==> 2NO You go through the same reasoning as the last problem we did. You didn't have pressure in that one but the rule on pressure is an increase in P shifts the equilibrium to the side with fewer mols gas. Since you have 2 mols on the left and 2 mols on...
February 25, 2015

chemistryYou should have told me you can get to that and we could have cut out half of the work. If you use atm for P, then R is 0.08206 L*atm/mol*K.
February 25, 2015

chemistrymols KClO3 = grams/molar mass = ? Using the coefficients in the balanced equation, convert mols KClO3 to mols O2. Now use PV = nRT at the conditions listed to calculate volume in L O2.
February 25, 2015

chemistryLe Chatelier's Principle tells us that if we do something to a system at equilibrium that the system will react so as to undo what we did to it. I will rewrite the equation like this, N2(g) + 3H2(g) ==> 2NH3(g) + heat. So if we decrease T, it will try to increase T. How...
February 25, 2015

chemistryBTW, just a reminder that THERE ARE NO UNITS in Kp, Kc, Ksp, Keq, etc. I know some profs use units (some because they don't know better) and if they call them provisional units that makes it ok, but technically, activities go in those constants and activities don't ...
February 25, 2015

ChemistryI'm not quite sure of how much "estimation" the author wants but this is what I would do. .......HSO4 ==> H^+ + SO4^2- I......0.15.....0......0 C.......-x......x......x E....0.15-x.....x......x K2 for H2SO4 = (x)(x)/(0.15-x) I would solve that exactly which ...
February 25, 2015

ChemistryI would classify HCl as a molecular compound. I also call HCl gas a polar covalent compound. It conducts electricity when placed in solution with H2O because it reacts with H2O to produce ions. HCl(g) + H2O(l) ==> H3O^+(aq) + Cl^-(aq)
February 25, 2015

chemistry#4 is not true; the others are true. I why #3 has that "according to the book". There are many posts I see and even a few texts that claim to show units for Keq, Ksp, Kc, Kp. etc. but in fact according to the book or not there are no units for these constants. The ...
February 25, 2015

ChemistryNeither sucrose nor paraffin in molten form will conduct electricity. Palmitic acid has a pKa of 4.78 which is almost the same as acetic acid so I would expect it to conduct but be a poor conductor typical of a weak acid.
February 25, 2015